Question

A car hits a tree at an estimated speed of 10 mi/hr on a 2% downgrade. If skid marks of 100 ft. are observed on dry pavement (F=0.33) followed by 200 ft. on an unpaved shoulder (F=0.28), what is the initial speed of the vehicle just before the pavement skid was begun?

Answer 1

Answer:

$v_{o} = 22.703\,\frac{m}{s}$ $\left(50.795\,\frac{m}{s}\right)$

Explanation:

The deceleration of the car on the dry pavement is found by the Newton's Law:

$\Sigma F = -\mu_{k,1}\cdot m\cdot g \cdot \cos \theta + m\cdot g \cdot \sin \theta = m\cdot a_{1}$

Where:

$a_{1} = (-\mu_{k,1}\cdot \cos \theta + \sin \theta)\cdot g$

$a_{1} = (-0.33\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$a_{1} = -3.040\,\frac{m}{s^{2}}$

Likewise, the deceleration of the car on the unpaved shoulder is:

$a_{2} = (-\mu_{k,2}\cdot \cos \theta + \sin \theta)\cdot g$

$a_{2} = (-0.28\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$a_{2} = -2.549\,\frac{m}{s^{2}}$

The speed just before the car entered the unpaved shoulder is:

$v_{o} = \sqrt{\left(4.469\,\frac{m}{s} \right)^{2}-2\cdot \left(-2.549\,\frac{m}{s^{2}} \right)\cdot (60.88\,m)}$

$v_{o} = 18.175\,\frac{m}{s}$

And, the speed just before the pavement skid was begun is:

$v_{o} = \sqrt{\left(18.175\,\frac{m}{s} \right)^{2}-2\cdot \left(-3.040\,\frac{m}{s^{2}} \right)\cdot (30.44\,m)}$

$v_{o} = 22.703\,\frac{m}{s}$ $\left(50.795\,\frac{m}{s}\right)$